23. The points $$ A(1,4), B(-2,0) $$ and $$ C(4,-2) $$ of a triangle are mapped onto $$ A^{1}(7,4) $$, $$ B^{2}(x, y) $$ and $$ \mathrm{C}^{1}(10,16) $$ by a transformation $$ \mathrm{N}=\left(\begin{array}{ll}a & b \ c & d\end{array}\right) $$. Find (i) Matrix N of the transformation (4mks) (ii) Coordinates of $$ \mathrm{B}^{1} $$ (2mks) (iii) $$ A^{I I} B^{I I} C^{I I} $$ are the image of $$ A^{1} B^{2} C^{1} $$ under transtormation represented by matrix $$ \mathrm{M}=\left(\begin{array}{cc} 2 & -1 \ 0 & 0 \end{array}\right) \text { Write down the co-ordinates of } \mathrm{A}^{\mathrm{ur}} \mathrm{~B}^{\mathrm{I}} \mathrm{C}^{\mathrm{II}} $$ (vi)A transformation $$ N $$ followed by $$ M $$ can be represented by a single transformation K . Determine K (2mks)

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**(i) Finding the matrix $$ N $$**

The transformation is given by  
$$
N=\begin{pmatrix} a & b \ c & d \end{pmatrix}
$$
which maps any point $$(x,y)$$ to 
$$
\left(a x+by,\; c x+dy\right).
$$
We are told that
$$
A(1,4)\to A^1(7,4) \quad \text{and} \quad C(4,-2)\to C^1(10,16).
$$

*For point $$ A(1,4) $$:*  
The image is given by  
$$
a\cdot 1 + b \cdot 4 = 7 \quad \Longrightarrow \quad a+4b=7 \quad (1),
$$
$$
c\cdot 1 + d \cdot 4 = 4 \quad \Longrightarrow \quad c+4d=4 \quad (2).
$$

*For point $$ C(4,-2) $$:*  
Its image gives  
$$
a\cdot 4 + b\cdot (-2)=10 \quad \Longrightarrow \quad 4a-2b =10 \quad (3),
$$
$$
c\cdot 4 + d\cdot (-2)=16 \quad \Longrightarrow \quad 4c-2d =16 \quad (4).
$$

**Solve for $$a$$ and $$b$$:**

From (1):  
$$
a=7-4b.
$$
Substitute into (3):
$$
4(7-4b)-2b=10 \quad \Longrightarrow \quad 28-16b-2b=10,
$$
$$
28-18b=10 \quad \Longrightarrow \quad 18b=18,
$$
$$
b=1.
$$
Then from $$a=7-4b$$:
$$
a=7-4=3.
$$

**Solve for $$c$$ and $$d$$:**

From (2):  
$$
c=4-4d.
$$
Substitute into (4):
$$
4(4-4d)-2d=16 \quad \Longrightarrow \quad 16-16d-2d=16,
$$
$$
16-18d=16 \quad \Longrightarrow \quad -18d=0,
$$
$$
d=0.
$$
Then from $$c=4-4d$$:
$$
c=4.
$$

Thus, the matrix is  
$$
N=\begin{pmatrix} 3 & 1 \ 4 & 0 \end{pmatrix}.
$$

---

**(ii) Finding the coordinates of $$ B^{1} $$**

Even though in the mapping statement the image of $$B$$ was labeled $$B^2(x,y)$$, we now determine the image (denoted here by $$B^1$$) of $$B(-2,0)$$ under the transformation $$N$$:

$$
B=(-2,0) \quad \Longrightarrow \quad N\begin{pmatrix} -2 \ 0 \end{pmatrix} =
\begin{pmatrix} 3\cdot (-2)+1\cdot 0 \ 4\cdot (-2)+0\cdot 0 \end{pmatrix} =
\begin{pmatrix} -6 \ -8 \end{pmatrix}.
$$

Thus,  
$$
B^1=(-6,-8).
$$

---

**(iii) Finding the coordinates of $$ A^{II} $$, $$ B^{II} $$ and $$ C^{II} $$**

The triangle $$ A^1B^2C^1 $$ is transformed by the matrix  
$$
M=\begin{pmatrix} 2 & -1 \ 0 & 0 \end{pmatrix}.
$$

The images are found by multiplying each point by $$M$$.

1. For $$ A^1(7,4) $$:
   $$
   M\begin{pmatrix} 7 \ 4 \end{pmatrix} =
   \begin{pmatrix} 2\cdot7-1\cdot4 \ 0 \end{pmatrix} =
   \begin{pmatrix} 14-4 \ 0 \end{pmatrix} =
   \begin{pmatrix} 10 \ 0 \end{pmatrix},
   $$
   so $$ A^{II}=(10,0) $$.

2. For $$ B^2 $$ (which is $$ B^1 $$ found above, i.e. $$ (-6,-8) $$):
   $$
   M\begin{pmatrix} -6 \ -8 \end{pmatrix} =
   \begin{pmatrix} 2\cdot(-6)-(-8) \ 0 \end{pmatrix} =
   \begin{pmatrix} -12+8 \ 0 \end{pmatrix} =
   \begin{pmatrix} -4 \ 0 \end{pmatrix},
   $$
   so $$ B^{II}=(-4,0) $$.

3. For $$ C^1(10,16) $$:
   $$
   M\begin{pmatrix} 10 \ 16 \end{pmatrix} =
   \begin{pmatrix} 2\cdot10-1\cdot16 \ 0 \end{pmatrix} =
   \begin{pmatrix} 20-16 \ 0 \end{pmatrix} =
   \begin{pmatrix} 4 \ 0 \end{pmatrix},
   $$
   so $$ C^{II}=(4,0) $$.

---

**(iv) Finding the single transformation $$ K $$ such that $$ N $$ followed by $$ M $$ is equivalent to $$ K $$**

Since the transformation $$ N $$ is applied first and then $$ M $$, the combined transformation is represented by  
$$
K = M \cdot N.
$$

Recall:
$$
M=\begin{pmatrix} 2 & -1 \ 0 & 0 \end{pmatrix} \quad \text{and} \quad N=\begin{pmatrix}3 & 1 \ 4 & 0 \end{pmatrix}.
$$

Compute the product:
$$
K=M\cdot N=\begin{pmatrix} 2 & -1 \ 0 & 0 \end{pmatrix}\begin{pmatrix} 3 & 1 \ 4 & 0 \end{pmatrix}.
$$

The entries of $$ K $$ are:

- First row, first column:
  $$
  2\cdot 3 + (-1)\cdot 4 = 6-4 = 2.
  $$
- First row, second column:
  $$
  2\cdot 1 + (-1)\cdot 0 = 2+0 = 2.
  $$
- Second row: both entries are $$ 0 $$ since the second row of $$M$$ is $$ (0,0) $$.

Thus,
$$
K=\begin{pmatrix} 2 & 2 \ 0 & 0 \end{pmatrix}.
$$
**(i) Matrix $$ N $$ of the transformation:** $$ N = \begin{pmatrix} 3 & 1 \ 4 & 0 \end{pmatrix} $$ **(ii) Coordinates of $$ B^{1} $$:** $$ B^{1} = (-6, -8) $$ **(iii) Coordinates of $$ A^{II} $$, $$ B^{II} $$, and $$ C^{II} $$:** $$ A^{II} = (10, 0), \quad B^{II} = (-4, 0), \quad C^{II} = (4, 0) $$ **(iv) Single transformation $$ K $$:** $$ K = \begin{pmatrix} 2 & 2 \ 0 & 0 \end{pmatrix} $$

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